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karlor · Aug 17, 2012, 07:45 AM

Is there an equation to see how high thrown objects go?

RickJ · Aug 17, 2012, 08:26 AM

Yes there is. It includes the force on the object, the weight of the object, the atmosphere, wind and many other things.

ebaines · Aug 17, 2012, 08:27 AM

The height the javelin reaches can be estimated if you know the amount of time the javelin is in the air:

$<br /> h =\frac{gt^2} 8<br />$

where $g$ = acceleration due to gravity (32.2 ft/s^2, or 9.8 m/s^2) and $t$ = time from release to hitting the ground. This formula assumes that the maximum height attained is "large" compared to the height above the ground when the javelin is released.

Alternatively if you know the distance the javelin is thrown and assume that the thrower launched the javelin at precisely 45 degrees from horizontal it's possible to make an estimate, but that assumption may be unrealistic. Ignoring the effects of air resistance the max height woud be 1/4 of the distance of the throw.

RickJ · Aug 17, 2012, 08:34 AM

If we don't know the time in the air, can we use this formula?:
h = (0.5*v^2)/g

karlor · Aug 20, 2012, 07:08 AM

Originally Posted by RickJ

If we don't know the time in the air, can we use this formula?:
h = (0.5*v^2)/g

How would that work?

ebaines · Aug 20, 2012, 08:42 AM

Originally Posted by RickJ

If we don't know the time in the air, can we use this formula?:
h = (0.5*v^2)/g

Yes this works, where v = the vertical component of the javelin's velocity at the instant it is released. Although it may be impractical to determine what this velocity is without specialized equipment.